Question

tim các số nguyên x , y biết : x+y = x.y+6

Answer 1

`Answer:`

\(x+y=x.y+6\)

\(\Leftrightarrow x+y-xy-6=0\)

\(\Leftrightarrow\left(x-xy\right)+y-6=0\)

\(\Leftrightarrow x\left(1-y\right)+y-1-5=0\)

\(\Leftrightarrow x\left(1-y\right)-\left(1-y\right)=5\)

\(\Leftrightarrow\left(x-1\right)\left(1-y\right)=5\)

Ta có: \(5=\left(-5\right).\left(-1\right)=\left(-1\right).\left(-5\right)=5.1=1.5\)

Ta có bảng sau:

x - 1	5	-5	1	-1
1 - y	1	-1	5	-5
x	6	-4	2	0
y	0	2	-4	6

Vậy `(x;y)\in{(6;0),(4;2),(2;-4),(0;6)}`

Answer 2

x+y=x.y+6 => x.y+6-x-y=0 => x.y-x+6-y=0 => x(y-1)+1-y+5 = 0 => x(y-1) -(y-1) = -5 =>(x-1)(y-1) = -5

Rồi tính tiếp ra các cặp (x;y) : (2;-4), (6;0), (0;6), (-4;2)